---
book: 13
number: 8
id: "XIII.8"
kind: "theorem"
uses: ["[[book-4/proposition-14]]", "[[book-1/proposition-4]]", "[[book-1/proposition-32]]", "[[book-3/proposition-28]]", "[[book-6/proposition-33]]", "[[book-1/proposition-6]]", "[[book-1/proposition-5]]", "[[book-6/proposition-4]]", "[[book-5/proposition-14]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.8"
license: "CC-BY-SA-4.0"
---

# XIII.8

*If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order*, *they cut one another in extreme and mean ratio*, *and their greater segments are equal to the side of the pentagon.*

## Proof

For in the equilateral and equiangular pentagon *ABCDE* let the straight lines *AC*, *BE*, cutting one another at the point *H*, subtend two angles taken in order, the angles at *A*, *B*; I say that each of them has been cut in extreme and mean ratio at the point *H*, and their greater segments are equal to the side of the pentagon.

For let the circle *ABCDE* be circumscribed about the pentagon *ABCDE*. [[book-4/proposition-14|IV. 14]]

Then, since the two straight lines *EA*, *AB* are equal to the two *AB*, *BC*, and they contain equal angles, therefore the base *BE* is equal to the base *AC*, the triangle *ABE* is equal to the triangle *ABC*, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [[book-1/proposition-4|I. 4]]

Therefore the angle *BAC* is equal to the angle *ABE*; therefore the angle *AHE* is double of the angle *BAH*. [[book-1/proposition-32|I. 32]]

But the angle *EAC* is also double of the angle *BAC*, inasmuch as the circumference *EDC* is also double of the circumference *CB*; [[book-3/proposition-28|III. 28]], [[book-6/proposition-33|VI. 33]] therefore the angle *HAE* is equal to the angle *AHE*; hence the straight line *HE* is also equal to *EA*, that is, to *AB*. [[book-1/proposition-6|I. 6]]

And, since the straight line *BA* is equal to *AE*, the angle *ABE* is also equal to the angle *AEB*. [[book-1/proposition-5|I. 5]]

But the angle *ABE* was proved equal to the angle *BAH*; therefore the angle *BEA* is also equal to the angle *BAH*.

And the angle *ABE* is common to the two triangles *ABE* and *ABH*; therefore the remaining angle *BAE* is equal to the remaining angle *AHB*; [[book-1/proposition-32|I. 32]] therefore the triangle *ABE* is equiangular with the triangle *ABH*; therefore, proportionally, as *EB* is to *BA*, so is *AB* to *BH*. [[book-6/proposition-4|VI. 4]]

But *BA* is equal to *EH*; therefore, as *BE* is to *EH*, so is *EH* to *HB*.

And *BE* is greater than *EH*; therefore *EH* is also greater than *HB*. [[book-5/proposition-14|V. 14]]

Therefore *BE* has been cut in extreme and mean ratio at *H*, and the greater segment *HE* is equal to the side of the pentagon.

Similarly we can prove that *AC* has also been cut in extreme and mean ratio at *H*, and its greater segment *CH* is equal to the side of the pentagon. Q. E. D.